Cho A(-3;6) , B(1;-2) và C(6;3) , có AD là đường phân giác góc A (D thuộc BC ) Tìm tọa độ D

Đáp án:

\(\left[ \begin{array}{l}
D\left( { – 39 + 30\sqrt 2 ;\,\, – 42 + 30\sqrt 2 } \right)\\
D\left( { – 39 – 30\sqrt 2 ;\,\, – 42 – 30\sqrt 2 } \right)
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
Phuong\,\,trinh\,\,\,duong\,\,thang\,\,\,BC:\\
\frac{{x – 1}}{{6 – 1}} = \frac{{y + 2}}{{3 + 2}}\\
\Leftrightarrow x – 1 = y + 2\\
\Leftrightarrow x – y – 3 = 0\\
D \in BC \Rightarrow D\left( {d;\,\,d – 3} \right).\\
\overrightarrow {AB} = \left( {4;\, – 8} \right) \Rightarrow AB = 4\sqrt 5 \\
\overrightarrow {AC} = \left( {9;\, – 3} \right) \Rightarrow AC = 3\sqrt {10} \\
\overrightarrow {DB} = \left( {d – 1;\,\,d – 1} \right) \Rightarrow BD = \sqrt {2{{\left( {d – 1} \right)}^2}} \\
\overrightarrow {DC} = \left( {d – 6;\,\,d – 6} \right) \Rightarrow DC = \sqrt {2{{\left( {d – 6} \right)}^2}} \\
Ap\,\,dung\,\,tinh\,\,\,chat\,\,tia\,\,\,phan\,\,giac\,\,\,cua\,\,\Delta \,\,\,ta\,\,co:\\
\frac{{AB}}{{AC}} = \frac{{BD}}{{DC}} \Leftrightarrow \frac{{4\sqrt 5 }}{{3\sqrt {10} }} = \frac{{\sqrt {2{{\left( {d – 1} \right)}^2}} }}{{\sqrt {2{{\left( {d – 6} \right)}^2}} }}\\
\Leftrightarrow \frac{{4\sqrt 5 }}{{3\sqrt {10} }} = \frac{{\left| {d – 1} \right|}}{{\left| {d – 6} \right|}}\\
\Leftrightarrow 4\sqrt 5 \left| {d – 6} \right| = 3\sqrt {10} \left| {d – 1} \right|\\
\Leftrightarrow 80{\left( {d – 6} \right)^2} = 90{\left( {d – 1} \right)^2}\\
\Leftrightarrow 8\left( {{d^2} – 12d + 36} \right) = 9\left( {{d^2} – 2d + 1} \right)\\
\Leftrightarrow {d^2} + 78d – 279 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
d = – 39 + 30\sqrt 2 \\
d = – 39 – 30\sqrt 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
D\left( { – 39 + 30\sqrt 2 ;\,\, – 42 + 30\sqrt 2 } \right)\\
D\left( { – 39 – 30\sqrt 2 ;\,\, – 42 – 30\sqrt 2 } \right)
\end{array} \right..
\end{array}\)